Self energy of non conducting sphere of radius ‘R’ carrying charge ‘Q’ is _____$\left[\mathrm{k}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\right]$

Uniformly charged sphere

Charge density $\mathrm{\rho }=\frac{3\mathrm{Q}}{4{\mathrm{\pi R}}^3}$

For a sphere of radius = x

charge contained (q) $=\frac{4}{3}\pi x^3\cdot \rho =\frac{\mathrm{Q}}{{\mathrm{R}}^3}\cdot {\mathrm{x}}^3$ 

Potential at that surface is $\mathrm{V}=\frac{1}{4\mathrm{\pi }{\in }_0}\cdot \frac{\mathrm{q}}{\mathrm{x}}$

charge needed to increase the radius from x to x + dx is

$\mathrm{dq}=\left(4{\mathrm{\pi x}}^2\cdot \mathrm{dx}\right)\cdot \mathrm{\rho }=\frac{3\mathrm{Q}}{{\mathrm{R}}^3}\cdot {\mathrm{x}}^2\cdot \mathrm{dx}$

work done in bringing the charge dq from infinity to the surface of radius x is dW = V

$\mathrm{dq}=\frac{3{\mathrm{q}}^2}{4{\mathrm{\pi ϵ}}_0\cdot {\mathrm{R}}^6}\cdot {\mathrm{x}}^4\mathrm{dx}$

Total work done in assembly the charged sphere of radius R is

$\begin{array}{l}\mathrm{W}={\int }_0^{\mathrm{R}} \frac{3{\mathrm{q}}^2}{4{\mathrm{\pi ϵ}}_0\cdot {\mathrm{R}}^6}\cdot {\mathrm{x}}^4\cdot \text{ due }\\ =\frac{3{\mathrm{Q}}^2}{20{\mathrm{\pi ϵ}}_0\cdot \mathrm{R}} \left[\text{ if }\frac{1}{4{\mathrm{\pi ϵ}}_0}=\mathrm{k}\right]\end{array}$